Bernoulli trials.

I am a surgeon. If a particular complication for a particular operation is 1%, what are the chances I will encounter this particular complication if I perform this particular operation on 13 patients?
Many thanks,
Sweetcaroline6

Last edited by mr fantastic; May 14th 2011 at 06:18 AM.
Reason: Re-titled.

What are the chances? 1?
How many can you expect to see? 13 * 0.01 = 0.13 - but don't let this persuade you to think it is zero.

The value 1% is created by observing a large population, probably several 1000 cases. You simply must not apply this to a single individual.

A specific patieint's history, condition, response, etc. is far more important than the 1%.

Even if 1% is exacty correct. Do the operation 100 times and tell them all that there will be no complication. Are you really okay with the one individual, the one family, that is now gravely disappointed in your knowledge or ability or that now needs to talk to your lawyer?

In my view, you need to look back on the patient, not forward, before you discuss risks. Each patient is likely already in the group that will have the complication, you just don't know it. The risk should be discussed very carefully with EACH patient so that 99 of them can be surprised.

Technically, p(0) = 0.99^13 = 0.8775 -- Probability that you will not see a case with the complication
p(1) = 13(0.99^12)*0.01 = 0.1152 -- Probability that you will see one such case.
p(2) = 78(0.99^11)*(0.01^2) = 0.0070 -- Probability that you will see TWO. Did you know this was not zero?
p(3 or more) = 1-p(2)-p(1)-p(0) = 0.0003 -- Another perhaps surprising non-zero probability.

I am a surgeon. If a particular complication for a particular operation is 1%, what are the chances I will encounter this particular complication if I perform this particular operation on 13 patients?
Many thanks,
Sweetcaroline6